Solve 8sqrt{2k-1}+3=11

$$8 \sqrt { 2 k - 1 } + 3 = 11$$

Solve for k

$$k=1$$

Steps for Solving Radical Equations

Subtract 3 from both sides of the equation.

$$8\sqrt{2k-1}+3-3=11-3$$

Subtracting 3 from itself leaves 0.

$$8\sqrt{2k-1}=11-3$$

Subtract 3 from 11.

$$8\sqrt{2k-1}=8$$

Divide both sides by 8.

$$\frac{8\sqrt{2k-1}}{8}=\frac{8}{8}$$

Dividing by 8 undoes the multiplication by 8.

$$\sqrt{2k-1}=\frac{8}{8}$$

Divide 8 by 8.

$$\sqrt{2k-1}=1$$

Square both sides of the equation.

$$2k-1=1$$

Add 1 to both sides of the equation.

$$2k-1-\left(-1\right)=1-\left(-1\right)$$

Subtracting -1 from itself leaves 0.

$$2k=1-\left(-1\right)$$

Subtract -1 from 1.

$$2k=2$$

Divide both sides by 2.

$$\frac{2k}{2}=\frac{2}{2}$$

Dividing by 2 undoes the multiplication by 2.

$$k=\frac{2}{2}$$

Divide 2 by 2.

$$k=1$$